Is (cos(θ)-sin(θ))^2=1-sin(2θ) ?

The answer to whether (cos(θ)-sin(θ))^2=1-sin(2θ) is True

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prove (cos(θ)-sin(θ))^2=1-sin(2θ)

prove

(cos (θ) - sin (θ))^{2} = 1 - sin (2 θ)

True

Solution steps

(cos (θ) - sin (θ))^{2} = 1 - sin (2 θ)

Manipulating left side

(cos (θ) - sin (θ))^{2}

Expand

(cos (θ) - sin (θ))^{2} : cos^{2} (θ) - 2 cos (θ) sin (θ) + sin^{2} (θ)

(cos (θ) - sin (θ))^{2}

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = cos (θ), b = sin (θ)

= cos^{2} (θ) - 2 cos (θ) sin (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ) - 2 cos (θ) sin (θ)

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ) - 2 cos (θ) sin (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= cos^{2} (θ) + sin^{2} (θ) - sin (2 θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

= 1 - sin (2 θ)

= 1 - sin (2 θ)

We showed that the two sides could take the same form

\Rightarrow True

(cos (x) - sin (x))^{2} = sec^{2} (x) - 2 tan (x)

\frac{2 sec ^{2} ( x ) - 2 tan ^{2} ( x )}{csc ( x )} = sin (2 x) sec (x)

sin^{2} (θ) - 2 = - 1 - cos^{2} (θ)

sin (2 α) = 2 sin (α) cos (α)

cos (θ + π) = - cos (θ)

Is (cos(θ)-sin(θ))^2=1-sin(2θ) ?
The answer to whether (cos(θ)-sin(θ))^2=1-sin(2θ) is True